A121149 -id:A121149 - cp's OEIS frontend

A284373 Number of distinct planar connected n-polyhexes having a minimal number of vertices.

1, 1, 1, 1, 1, 3, 1, 1, 4, 1, 2, 1, 2, 1, 3, 1, 6, 3, 1, 1, 1, 7, 4, 1, 4, 2, 1, 3, 2, 1, 4, 3, 1, 9, 6, 3, 1, 2, 1, 1, 10, 7, 4, 1, 5, 4, 2, 1, 4, 3, 2, 1, 6, 4, 3, 1, 12, 9, 6, 3, 1, 2, 2, 1, 1, 13, 10, 7, 4, 1, 7, 5, 4, 2, 1, 5, 4, 3, 2, 1, 7, 6, 4, 3, 1, 15, 12, 9, 6

Offset: 1

Luca Petrone, Mar 25 2017

nonn

Luca Petrone, Table of n, a(n) for n = 1..386
Eric Weisstein's World of Mathematics, Polyhex.

Cf. A121149.

polyhexeQ[{{_Integer, _Integer} .. }] := True
polyhexeQ[_] := False
rot[p_?polyhexeQ] := {-Last[#], Plus @@ #} & /@ p
ref[p_?polyhexeQ] := {-Plus @@ #, Last[#]} & /@ p
cyclic[p_] := Module[{i = p, ans = {p}},
  While[(i = rot[i]) != p, AppendTo[ans, i]]; ans]
dihedral[p_?polyhexeQ] := Flatten[{#, ref[#]} & /@ cyclic[p], 1]
canonical[p_?polyhexeQ] :=
Sort[Map[(# - {Min[First /@ p], Min[Last /@ p]}) &, p]] allPieces[p_] := Union[canonical /@ dihedral[p]]
polyhexes[1] := {{{0, 0}}}
polyhexes[n_] :=
polyhexes[n] =
  Module[{f, a, b, fig, ans = {}},
   fig = Map[(f = #; Map[({a, b} = #; {f, {a - 1, b - 1}, f, {a + 1, b - 2}, f, {a + 2, b - 1}, f, {a + 1, b + 1}, f, {a - 1, b + 2}, f, {a - 2, b + 1}}) &, f]) &, polyhexes[n - 1]];
   fig = Partition[Partition[Flatten[fig], 2], n];
   f = Union[canonical /@ Select[Union /@ fig, Length[#] == n &]];
   While[f != {},
    ans = {ans, First[f]};
    f = Complement[f, allPieces[First[f]]]];
   Partition[Partition[Flatten[ans], 2], n]]
coord[z_] := {Re[#], Im[#]} & /@ z
atoms[p_?polyhexeQ] := Module[{a, b, v, t, u = E^(Pi I/3)}, {{a, b} = #; v = a + b u; coord[{v, v + 1, v + 1 + u, v + 2 u, v + 2 u - 1, v + u - 1}]} & /@ p]
A = {};
n = 1;
While[n <= 386,
polyhexes[n];
polyhexes[n] = Part[polyhexes[n], #] & /@ Ordering[Length[Tally[Flatten[atoms[#], 2]]] &  /@ polyhexes[n],     BinCounts[#, {Min[#], Min[#] + 1}][[1]] & [Length[Tally[Flatten[atoms[#], 2]]] &  /@ polyhexes[n]]];
A = Flatten[{A, Length[#]}] & [Length[Tally[Flatten[atoms[#], 2]]] &  /@ polyhexes[n]];
Print[A[[n]]];
n++;]
(* Luca Petrone, Mar 25 2017, based on a program by Jaime Rangel-Mondragón *)

A182618 Number of new grid points that are covered by the toothpicks added at n-th-stage to the toothpick spiral of A182617.

Original entry on oeis.org

6, 4, 3, 3, 3, 3, 2, 3, 3, 2, 3, 2, 3, 2, 3, 2, 3, 2, 2, 3, 2, 3, 2, 2, 3, 2, 2, 3, 2, 2, 3, 2, 2, 3, 2, 2, 2, 3, 2, 2, 3, 2, 2, 2, 3, 2, 2, 2, 3, 2, 2, 2, 3, 2, 2, 2, 3, 2, 2, 2, 2, 3, 2, 2, 2, 3, 2, 2, 2, 2, 3, 2, 2, 2, 2, 3

Offset: 1

Omar E. Pol, Dec 12 2010

In the toothpick spiral the toothpicks are connected by their endpoints. See A182617 for more information.
Attempt at an explanation, R. J. Mathar, Dec 13 2010: (Start)
In the hexagonal grid, we can pick any of the hexagons as a center, and then define a ring of 6 first neighbors (hexagons adjacent to the center), then define a ring of 12 second neighbors (hexagons adjacent to any of the first ring) and so on. The current sequence describes a self-avoiding walk which starts in a spiral around the center hexagon, which covers 5 edges. The walk then takes one step to reach the rim of the first ring and travels once around this ring until it reaches a point where self-avoidance stops it. It then takes one step to reach the rim of the second ring and walks around that one, etc. Imagine that on each edge we place a toothpick if it's on the path, and interrupt counting the total number of toothpicks each time one of the hexagons has six vertices covered. The first differences of these intermediate totals define the sequence. (End)

			At stage 1, starting from a node on the hexagonal net, we place 5 toothpicks on 5 edges of the first hexagon, so a(1)= 6 because there are 6 grid points that are covered by the toothpicks.
At stage 2, starting from the last exposed endpoints, we place 4 toothpicks on the edges of the second hexagon, so a(2)=4 because there are new 4 grid points that are covered by the toothpicks.
At stage 3, starting from the last exposed endpoints we place 3 toothpicks on the edges of the third hexagon, so a(3)=3 because there are new 3 grid points covered. Etc.
If written as a triangle, begins:
6,
4,3,3,3,3,2,
3,3,2,3,2,3,2,3,2,3,2,2,
3,2,3,2,2,3,2,2,3,2,2,3,2,2,3,2,2,2,
3,2,2,3,2,2,2,3,2,2,2,3,2,2,2,3,2,2,2,3,2,2,2,2,
3,2,2,2,3,2,2,2,2,3,2,2,2,2,3,2,2,2,2,3,2,2,2,2,3,2,2,2,2,2

David Applegate, Omar E. Pol and N. J. A. Sloane, The Toothpick Sequence and Other Sequences from Cellular Automata, Congressus Numerantium, Vol. 206 (2010), 157-191. [There is a typo in Theorem 6: (13) should read u(n) = 4.3^(wt(n-1)-1) for n >= 2.]
N. J. A. Sloane, Catalog of Toothpick and Cellular Automata Sequences in the OEIS
Index entries for sequences related to toothpick sequences

Cf. A121149, A139250, A139251, A182617, A182619, A182632, A182840.

Row n has A008458(n-1) terms. Row sums give A017593.

A141135 Minimal number of unit edges required to construct n regular pentagons when allowing edge-sharing.

Original entry on oeis.org

5, 9, 13, 17, 21, 24, 28, 32, 36, 39, 43, 47, 50, 54, 58, 61, 65, 69, 72, 76, 80, 83, 87, 90, 94, 98, 101, 105, 109, 112

Offset: 1

Ralph H. Buchholz (teufel_pi(AT)yahoo.com), Jun 08 2008

			a(6) = 24 since the first pentagon requires 5 edges, the 2nd, 3rd, 4th and 5th pentagons require an additional 4 edges each and the 6th pentagon requires 3 edges since it can share 2 edges (if one tiles via a 6-cycle). Thus 24 = 5 + 4 + 4 + 4 + 4 + 3.

Ralph H. Buchholz, Spiral polygon series, preprint 1985 SMJ 31, School Mathematics Journal, 1995.
Ralph H. Buchholz and Warwick de Launey, Edge minimization, June 1996, (revised June 2008).
Ralph H. Buchholz and Warwick de Launey, An edge minimization problem for regular polygons, The Electronic Journal of Combinatorics, Volume 16, Issue 1 (2009), #R90.

Cf. equilateral triangles A137228, squares A078633, regular hexagons A135708.

Cf. A121149.

Conjectures from Colin Barker, Apr 05 2019: (Start)
G.f.: x*(5 + 4*x + 4*x^2 - x^3 - x^5 + x^8 - x^9) / ((1 - x)^2*(1 + x + x^2)).
a(n) = a(n-1) + a(n-3) - a(n-4) for n>10.
(End)
Conjecture: if n is a term in A121149, a(n) = a(n-1) + 3, otherwise a(n) = a(n-1) + 4. - Jinyuan Wang, Apr 05 2019

a(21)-a(30) from Jinyuan Wang, Apr 05 2019

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A284373 Number of distinct planar connected n-polyhexes having a minimal number of vertices.

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A182618 Number of new grid points that are covered by the toothpicks added at n-th-stage to the toothpick spiral of A182617.

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A141135 Minimal number of unit edges required to construct n regular pentagons when allowing edge-sharing.

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A121151 Minimal number of vertices in an n-polytrimino (or n-polyiamond).

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